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Search: id:A099514
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| A099514 |
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Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + z + 2*z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2. |
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+0
2
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| 1, 1, 1, 1, 2, 2, 1, 3, 5, 4, 1, 4, 9, 13, 4, 1, 5, 14, 28, 18, 12, 1, 6, 20, 50, 49, 56, 8, 1, 7, 27, 80, 105, 161, 56, 32, 1, 8, 35, 119, 195, 366, 210, 200, 16, 1, 9, 44, 168, 329, 72, 1, 581, 732, 160, 80, 1, 10, 54, 228, 518, 1288, 1337, 2045, 780, 640, 32, 1, 11, 65, 300
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Row sums form A099515. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).
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FORMULA
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G.f.: (1-x+x*y-2*x^2*y^2)/((1-x)^2-4*x^2*y^2+3*x^3*y^2+4*x^4*y^4).
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EXAMPLE
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Rows begin:
[1],
[1,1],
[1,2,2],
[1,3,5,4],
[1,4,9,13,4],
[1,5,14,28,18,12],
[1,6,20,50,49,56,8],
[1,7,27,80,105,161,56,32],
[1,8,35,119,195,366,210,200,16],
[1,9,44,168,329,721,581,732,160,80],...
and can be derived from coefficients of (1+z+2*z^2)^n:
[1],
[1,1,2],
[1,2,5,4,4],
[1,3,9,13,18,12,8],
[1,4,14,28,49,56,56,32,16],
[1,5,20,50,105,161,210,200,160,80,32],...
by shifting each column k down by [k/2] rows.
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PROGRAM
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(PARI) T(n, k)=if(n<k|k<0, 0, polcoeff((1+z+2*z^2+z*O(z^k))^(n-k\2), k, z))
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CROSSREFS
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Cf. A099509, A099510, A099512, A099515.
Adjacent sequences: A099511 A099512 A099513 this_sequence A099515 A099516 A099517
Sequence in context: A079956 A140717 A026300 this_sequence A139687 A064581 A064580
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Oct 20 2004
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